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Comparative Problems

Comparative Problems

Comparative problems involve calculation of something in terms of something else that we know. For example, if you weigh \(\text{490}\) \(\text{N}\) on Earth and the gravitational acceleration on Venus is \(\text{0.903}\) that of the gravitational acceleration on the Earth, then you would weigh \(0.903\times \text{490} \text{N}=\text{442.5} \text{N}\) on Venus.

Method for Answering Comparative Problems

  • Write out equations and calculate all quantities for the given situation

  • Write out all relationships between variable from first andsecond case

  • Write out second case

  • Substitute all first case variables into second case

  • Write second case in terms of first case

Example: Comparative Problem

Question

A man has a mass of \(\text{70}\) \(\text{kg}\). The planet Zirgon is the same size as the Earth but has twice the mass of the Earth. What would the man weigh on Zirgon, if the gravitational acceleration on Earth is \(\text{9.8}\) \(\text{m·s$^{-2}$}\)?

Solution

Step 1: Determine what information has been given

The following has been provided:

  • the mass of the man, m
  • the mass of the planet Zirgon (\({m}_{Z}\)) in terms of the mass of the Earth (\({M}_{\text{Earth}}\)), \({m}_{Z}=2{M}_{\text{Earth}}\)
  • the radius of the planet Zirgon (\({r}_{Z}\)) in terms of the radius of the Earth (\({r}_{E}\)), \({r}_{Z}={r}_{\text{Earth}}\)

Step 2: Determine how to approach the problem

We are required to determine the man’s weight on Zirgon (\({w}_{Z}\)). We can do this by using:

\[{F}_{g}=mg=G\frac{{m}_{1}·{m}_{2}}{{d}^{2}}\]

to calculate the weight of the man on Earth and then use this value to determine the weight of the man on Zirgon.

Step 3: Situation on Earth

\begin{align*} {F}_{\text{Earth}}& = m{g}_{E}=G\frac{{M}_{\text{Earth}}·m}{{d}_{E}^{2}} \\ & = \left(\text{70}\text{ kg}\right)\left(\text{9.8}\text{ m·s$^{-2}$}\right) \\ & = \text{686}\text{ N} \end{align*}

Step 4: Situation on Zirgon in terms of situation on Earth

Write the equation for the gravitational force on Zirgon and then substitute the values for \({m}_{Z}\) and \({r}_{Z}\), in terms of the values for the Earth.

\begin{align*} {w}_{Z}=m{g}_{Z}& = G\frac{{m}_{Z}·m}{{r}_{Z}^{2}} \\ & = G\frac{2{M}_{\text{Earth}}·m}{{r}_{\text{Earth}}^{2}} \\ & = 2\left(G\frac{{M}_{\text{Earth}}·m}{{r}_{E}^{2}}\right) \\ & = 2{F}_{\text{Earth}} \\ & = 2\left(\text{686} \text{N}\right) \\ & = 1372 \text{N} \end{align*}

Step 5: Quote the final answer

The man weighs \(\text{1 372}\) \(\text{N}\) on Zirgon.

Example: Comparative Problem

Question

A man has a mass of \(\text{70}\) \(\text{kg}\). On the planet Beeble how much will he weigh if Beeble has a mass half of that of the Earth and a radius one quarter that of the Earth. Gravitational acceleration on Earth is \(\text{9.8}\) \(\text{m·s$^{-2}$}\).

Solution

Step 1: Determine what information has been given

The following has been provided:

  • the mass of the man on Earth, m
  • the mass of the planet Beeble (\({m}_{B}\)) in terms of the mass of the Earth (\(M_{Earth}\)),\(m_B=\frac{\text{1}}{\text{2}}M_{Earth}\)
  • the radius of the planet Beeble (\({r}_{B}\)) in terms of the radius of the Earth (\({r}_{E}\)), \({r}_{B}=\frac{\text{1}}{\text{4}}{r}_{E}\)

Step 2: Determine how to approach the problem

We are required to determine the man’s weight on Beeble (\({w}_{B}\)). We can do this by using: \begin{align*} F_g=mg=G\frac{m_1m_2}{d^2} \end{align*}

to calculate the weight of the man on Earth and then use this value to determine the weight of the man on Beeble.

Step 3: Situation on Earth

\begin{align*} F_{Earth}&= mg_{Earth} = G \frac{M_{Earth}}{r_E^2} \\ &= (\text{70})(\text{9.8}) \\ &=\text{686}\text{ N} \end{align*}

Step 4: Situation on Beeble in terms of situation on Earth

Write the equation for the gravitational force on Beeble and then substitute the values for \({m}_{B}\) and \({r}_{B}\), in terms of the values for the Earth.

\begin{align*} F_{Beeble}&= mg_{Beeble} = G \frac{M_{Beeble}}{r_B^2} \\ &= G \frac{\frac{\text{1}}{\text{2}}M_{Earth}}{\frac{\text{1}}{\text{4}}r_E^2} \\ & = \text{8}\left(G \frac{M_{Earth}}{r_E^2} \right) \\ &= 8 (\text{686}) \\ &=\text{5 488}\text{ N} \end{align*}

Step 5: Quote the final answer

The man weighs \(\text{5 488}\) \(\text{N}\) on Beeble.

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